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You should use 5 degrees here instead of 10. It's obvious if you draw a picture of the situation with 2 parallel lines representing the mentioned lines of longitudes, like on a mercator chart, with straight rhumb line between the points and a downwards pointing great circle line. The point where the g.c. is most south is in the middle between 20W and 30W, so 5 degrees longitude.

Even then I am not sure how accurate this answer is in the real world..

Back in the days we didn''t have this q. in the database. (pipe-smoking smiley)

i guess i've figured it out with some help from another forum. lets name the starting position A and the intermediate position on 25 degree W as B. when we join points A and B to form a triangle for calculation purpose, it divides conversion angle into two equal parts of 2.16 degress each. now solving the sum in the same manner: tan 2.16= X (unknown)/150 X=5.65nm X/60 = 0 degrees 5 minutes hence position B= 60 degrees + 5 minutes
=60 degrees 5 minutes

We can now draw a triangle with a horizontal side 60S 30W to 60S 25W. The length of this line will be 150 nm.

The second side is a vertical line from 60S 25W down to the unknown position south of 25W.

The third line is the hypotenuse form 60S 30W to the unknown position south of 25W.

The internal angle between the hypotenuse and the horizontal side is the conversion angle based on going from 60S 30W to 25W.

This angle = 0.5 x 5 degrees x Sin 60 = 2.165 degrees.

The tangent of this angle is the vertical side divided by the horizontal side.

Tan 2.165 = Ch lat / 150 nm

Rearranging this gives Ch lat = Tan 2.165 x 150 = 5.67 degrees.

This means that the change of latitude between 60S 30W and the unknown position south of 25W is 5.67 degrees. This makes the new latitude 60 degrees 5.67 minutes south. The closest option to this is 60°06’S

Sorry Anders my previous post did not address this part of your question.

Quote:

Most of the explanations I have seen for this question seem intent on halving the conversion angle, or rather getting the conversion angle from the ch.long between 030°W and 025°W, to get 2.2°-ish, and obviously that would present an answer more consistent with the correct one. But why is this done? I don’t see how that’s relevant as it would imply that we are flying from 60°N 030°W to 60°N 025°W and then on to 60°N 020°W, in which case I would fully agree with the answer, but that would mean that 025°W was an additional waypoint… Unless I'm missing something else.

To understand why they do this we need to sketch the whole picture. Draw a horizontal straight line to represent the 300 nm rhumb line track from 60S 30W to 60S 20W.

Now draw a shallow arc looping down between the two ends of the rhumb line track. This represents the great circle track.

Midway between the two ends of the rhumb line draw a vertical line down to the great circle arc. This vertical line is 150 nm from each end of the rhumb line track. The length of this vertical line represents the maximum change of latitude between the two tracks.

Now draw a sloping straight line from each end of the rhumb line to the lower end of the vertical line.

Using the conversion angle equation we can now calculate the angle between each end of the great circle arc and the ends of the rhumb line. This is 0.5 x 10 degrees x Sin 60S = 4.33 degrees. Write 4.33 in each of these angles.

Now let’s look at what happens when we fly from 60S 30W to the midway position south of 25W. We are initially tracking 094.33 degrees, but our track is continuously turning to the north, such that we are tracking 090 when we reach our most southerly point.

During this first half of the trip we have reduced our track direction by 4.33 degrees from 094.33 to 090. This means that our mean track was 092.165 degree. This mean track is represented by the straight line from our starting point to our most southerly point. So the internal angles in the triangles at each end of our track are 2.165 degrees. This is half of the conversion angle.

So in solving this type of problem it is easier to go straight for ½ the conversion angle and work out the solution from there.

Question: A turbojet aeroplane has a planned take-off mass of 190 000 kg; the cargo load is distributed as follows: cargo 1: 3 000 kg; (3.50 m from reference point) cargo 4: 7 000 kg. (20.39 m from reference point) Distance from reference point to leading edge: 14m Length of MAC = 4.6m. Once the cargo loading is completed, the crew is informed that the centre of gravity at take-off is located at 38 % MAC (Mean Aerodynamic Cord) which is beyond the limits. The captain decides then to redistribute part of the cargo load between cargo 1 and cargo 4 in order to obtain a new centre of gravity location at 31 % MAC. Following the transfer operation, the new load distribution is: cargo 1: 6 000 kg; cargo 4: 4 000 kg

This means that 3622 kg of cargo must be moved from hold 4 to hold 1 to move the C of G from 38% MAC to 31% MAC.

But the final line of your statement of the questions states that “ Following the transfer operation, the new load distribution is: cargo 1: 6 000 kg; cargo 4: 4 000 kg”. If this is correct then only 3000 kg has been moved, so the new C of G will not be at 31% MAC.

IF you didn't take into account the MRJT from CAP696/7 and just worked with the given figures then 13092 is the maximum traffic load, but not the clearest question ever produced. Would be better saying "an aircraft" with no reference to MRJT.